设x1、x2是方程x^2+(2k+1)x+k^2-2=0的两根x1+x2=2k+1x1x2=k^2-2(x1+x2)^2=(2k+1)^2x1^2+2x1x2+x2^2=4k^2+4k+1x1^2+x2^2=4k^2+4k+1-2x1x2x1^2+x2^2=4k^2+4k+1-2(k^2-2)x1^2+x2^2=2k^2+4k+5方程x^2+(2k+1)x+k^2-2=0的...